How do we factor the expression $a^2(b^3 - c^3) + b^2(c^3 - a^3) + c^2(a^3 - b^3)$?

Question

This a problem I came across in a high school book of the chapter polynomial. While solving I was sure that this has something to do with $(a-b)$ is a factor of $(a^3 - b^3)$. But after this I am stuck and cannot factorise the polynomial any further. If anyone knows the answer please share it. Thankyou

Answer 1

It is equal to $$(a-b)(b-c)(c-a)(ab+bc+ca)$$

Method: the cubic equation with $a$ as variable has 3 roots: $b,c$ and $-\frac{bc}{b+c}$

Answer 2

It is well known that $a - b$ divides $a^3 - b^3$. We can divide it out $(a^3 - b^3)/(a - b) = a^2 + ab + b^2$.

This leads us to ask whether we can also divide $a-b$ out as a factor of $a^2(b^3 - c^3) + b^2(c^3 - a^3)$. Consider this expression to be a polynomial in the variable $b$, if it vanishes at $b=a$ then $a-b$ will be a factor.

So we check $a^2(a^3 - c^3) + a^2(c^3 - a^3) = a^5 - a^2 c^3 + a^2 c^3 - a^5 = 0$.

The same logic applies for $b-c$ and $c-a$ so we can perform long division to find $$(a-b)(b-c)(c-a)(ab + bc + ac)$$

Answer 3

HINT

Here it is a trick which may help you now and then:

\begin{align*} c^{3} - a^{3} = (c^{3} - b^{3}) + (b^{3} - a^{3}) \end{align*}

Can you take it from here?

Answer 4

Since others have posted solutions, I may as well post mine. The observation that setting any pair of variable equal gives zero gives a factor $(a-b)(b-c)(c-a)$ (the sign does not matter at this stage). Note also the stronger property that swapping any two variables in the original changes the sign, so it is antisymmetric.

We therefore have $(a-b)(b-c)(c-a)X$ where $X$ is an unknown quadratic factor. Since the identified factor is antisymmetric, $X$ should be symmetric. Setting $a=0$ we get $$-bc(b-c)X_{a=0}=b^2c^2(c-b)$$ whence $X_{a=0}=bc$ and by symmetry $X=ab+bc+ca$. Then once you have the answer it can be checked.

Answer 5

How do we factor the expression $a^2(b^3 - c^3) + b^2(c^3 - a^3) + c^2(a^3 - b^3)$?

First, the condition for this expression is $a \ne b$, $b \ne c$, and $c \ne a$. In any event of $a = b$, or $b =e c$, or $c = a$, he expression becomes zero.

The difference of two cubes can be expressed as: $$x^3 - y^3 = (x - y)(x^2 + xy + y^2) \tag 1$$

Thus, $$a^2(b^3 - c^3) = a^2(b - c)(b^2 + bc + c^2) \tag2$$ and $$b^2(c^3 - a^3) + c^2(a^3 - b^3) = b^2c^3 - b^2a^3 + c^2a^3 - c^2b^3 = -a^3(b^2 - c^2) - b^2c^2(b - c) \\ = -a^3(b - c)(b + c) - b^2c^2(b - c) = -(b - c)(a^3b + a^3c - b^3c^2 - b^2c^3) \tag3$$

From the equations $(2)$ and $(3)$:

$$a^2(b^3 - c^3) + b^2(c^3 - a^3) + c^2(a^3 - b^3)\\ = a^2(b - c)(b^2 + bc + c^2) - (b - c)(a^3b + a^3c - b^3c^2 - b^2c^3) \\ = (b - c)(a^2b^2 + a^2bc + a^2c^2 - a^3b - a^3c + b^3c^2 + b^2c^3) \tag4$$

Therefore, $(b - c)$ is a factor of $a^2(b^3 - c^3) + b^2(c^3 - a^3) + c^2(a^3 - b^3)$. Similarly, you can prove that $(c - a)$ and $(a - b)$ are factors of the same:

$$a^2(b^3 - c^3) + b^2(c^3 - a^3) + c^2(a^3 - b^3) = (a-b)(b-c)(c-a)(\text{FF})$$

To find the $\text{FF}$ (forth factor, which should be a symmetric factor), you can either perform long division or use apply separately the conditions $a = 0$, $b = 0$, and $c = 0$. For instance, when $a = 0$:

$$a^2(b^3 - c^3) + b^2(c^3 - a^3) + c^2(a^3 - b^3) = b^2c^3 - c^2b^3 = (a-b)(b-c)(c-a)(\text{FF})$$ $$b^2c^3 - c^2b^3 = bc(b - c) (\text{FF})_{a = 0}$$ Since, $bc \ne 0$ and $b \ne c$, $\text{FF}_{a = 0} = bc$. Similarly, you can prove $\text{FF}_{b = 0} = ca$ and $\text{FF}_{c = 0} = ab$. Thus,

$$a^2(b^3 - c^3) + b^2(c^3 - a^3) + c^2(a^3 - b^3) = (a-b)(b-c)(c-a)(ab + bc + ca)$$